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Find the minimum value of 𝑥^2 − 4𝑥 − 12.
-16


i) (53/4) -5(x-(3/2))²
ii) x=3/2
( 3/2 , 53/4 )


bruh do NOT let the 5 marks and big words scare u its so easy


ONLYYYY do iii (i only have the answer for that)
i didn’t really understand what bisect mean so it was confusing but now it makes sense:
When you bisect a line segment, you cut it into two equal lengths, so you want to find the midpoint




super easy just kinda forgot, it’s a disguised quadratic, use your own method not the one that is here, but just look at the answer




Remember, this isn’t a two‑number answer — it’s a range of values for k.
The discriminant becomes a quadratic expression in k, and a quadratic is positive only between its two roots.
So the original quadratic has real roots only when k lies between those two values.
but if you got it right.. well done! ;p



What does it mean if e.g. f(x) = -6 has no real solution
It doesnt intersect the curve

bruh easy question too how did i get it wrong, this why you dont rush!


What i did wrong was i made it so y=x/2 , but its actually y=(x/2) + 1 , remember 2/2 is 1 , it doesnt cancel out into 0




I was on this question for a while, I knewww that I knew how to do it but I just didn’t know where I was going wrong, turns out I factorised x²+k²x² like (2k²)x²not like (1 + k²)x² how I was meant to so yh


just do b
the centre is (4,-5) and radius is √40
How do we know which is correct?
We use the condition: “the gradient of AQ is positive.”


ez 6 marks




btw we letting u= -4-k






